Re: EXPERIMENTS THAT REFUTE RELATIVITY

"Dr. Henri Wilson" <HW@....> wrote in message
news:68a0n3t5mcaumqrh60po29crbu8ernebe1@xxxxxxxxxx
On Mon, 24 Dec 2007 00:28:51 +1100, "Jeckyl" <noone@xxxxxxxxxxx> wrote:

"Dr. Henri Wilson" <HW@....> wrote in message
news:fasrm398mudvoln8i4is4np5akhf2krg3r@xxxxxxxxxx
On Sat, 22 Dec 2007 21:30:57 -0600, Tom Roberts
<tjroberts137@xxxxxxxxxxxxx>
wrote:

Dr. Henri Wilson wrote:
All that matters is the phase difference at the detector.

RIGHT!!! so why don't you follow your own advice?

According to emission theory, the two rays travel for the same time
duration
but their frequencies are doppler shifted in the nonrotating frame.

Look up at your previous statement (quoted above). It makes your
statement here completely irrelevant, because the detector is not at
rest in the nonrotating frame. The "Doppler shifts in the nonrotating
frame" cancel at the detector, and the two rays have the same frequency
AS SEEN BY THE DETECTOR.

Tom, DURING the time light is in transit around the ring is it DOPPLER
SHIFTED
IN THE NONROTATING FRAME.

It is doppler shifted in lots of different frames .. that doesn't really
matter

Of course it isn't doppler shifted wrt the detector.

So the waves leave the source at the same frequency and speed relative to
the source, they arrive at the detector at the same frequency and speed
realtive to the detector, they are in transit for the same time .. there
is
no Sagnac effect predicted.

Study the equations and point out where they are wrong or SHUT THE ***
UP!

They describe the phase difference at the initial source position in the
non-rotating frame.

As the wave are the same frequency and have travelled for the same time, by
the time they get to the moving detector (you have to Doppler shift to get
the frequency as the detector will 'see' it), that means the waves must
still be in phase. its very simple

Look at your train example.

Sagnac is similar to this.

There are three trains on a track .. A, B, C all moving at the same speed.
A fires a machine gun forward at B, C fires a machine gun backward at B.
The same rate of bullets arrives at B from both trains (because they are
comoving.

Now .. to make it more like Sagnac .. put the trains on a circular track,
and place some 'mirrors' for the bullets to bounce off that also move with
the train. Now, you don't need A and C to be separate trains .. they can
just be a single train on the opposite side of the circular track from train
B. Train A can shoot bullets forward and backward .. and both bullets
arrive at train B at the same time at the same rate.

Extend this example further, and you have a single train A shooting bullets
that bounce around the mirrors end up back at train A again. Now you have
Sagnac. You see that the bullets get fired from the machine guns in both
directions at the same rate and end up back at train A again at the same
rate and in the same time.

Its just blatantly obvious that there is no Sagnac effect in a ballistic
theory.


.

Quantcast